At the press conference this morning, the Event Horizon Telescope team didn't say much about Sagittarius A*, which was the target many of us have been waiting for.

Is there any explanation anywhere for this omission?


2 Answers 2


There was a mention of Sagittarius A* during the Q+A portion of the press conference; the team indicated that they hope to produce an image sometime in the future (although they were careful to make no promises, and they're not assuming they'll be successful).

That said, I'm not wholly surprised that we ended up seeing M87, rather than Sgr A*, for a couple reasons which the team mentions in their first paper:

  • As Glorfindel said, Sgr A*'s event horizon is much smaller, meaning matter orbiting the black hole has a shorter orbital period. This contributes to variability on the timescale of minutes. The observations of M87 took place over the course of a week - roughly the timescale over which that target varies, meaning the source should not change significantly over that time.
  • Second - and this is the reason I've seen cited more often - Sgr A* lies in the center of our galaxy, and so thick clouds of gas and dust lie between it and us. That results in scattering, which is a problem. There are ways to mitigate this, of course, and the team has spent a long time on this, but it's simpler to just look at the black hole that doesn't have that problem in the first place. That's why M87's black hole is an attractive target.

Neither of these are impossible hurdles to overcome, but they're certainly very real difficulties that can't be ignored.

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    $\begingroup$ I saw multiple reports before hand that it was expected that they'd release images of both. Did that turn out to be nothing more than speculation? $\endgroup$ Commented Apr 12, 2019 at 4:23
  • $\begingroup$ @curiousdannii I'd gotten definite information only the day before saying that it was going to be just M87, but I hadn't heard anything but rumors before that. Out of curiosity, where did you hear the reports? $\endgroup$
    – HDE 226868
    Commented Apr 16, 2019 at 14:54
  • $\begingroup$ mostly youtube videos I think. $\endgroup$ Commented Apr 16, 2019 at 23:02
  • $\begingroup$ The EHT observations are at mm wavelengths; the clouds of gas and dust between us and the galactic center are basically transparent to those wavelengths, so that's not really an issue. (It's an issue for optical and near-infrared observations, of course.) $\endgroup$ Commented Apr 18, 2019 at 9:02

I've found an explanation in Dutch here by Heino Falcke, one of the EHT founders. Translation:

Hard to photograph

It was easiest to take a picture of M87. "It is very difficult to photograph the black hole in our Milky Way, because the material around it moves very fast: the vortex rotates around its axis in 20 minutes. Compare it to a toddler who has to sit still for hours to be photographed: that's not possible. With M87, the matter revolves around the hole in two days, so it's easier to photograph", says Falcke.

(The original text is as follows:)

Lastig te fotograferen

Het lukte het beste om een foto te maken van M87. "Het is heel lastig om het zwarte gat in onze Melkweg op de foto te zetten, doordat de materie daaromheen heel snel beweegt: de draaikolk draait in 20 minuten om zijn as. Vergelijk het met een kleuter die urenlang stil moet zitten om op de foto te gaan: dat gaat niet. Bij M87 draait de materie in twee dagen om het gat heen, dus dat is makkelijker te fotograferen", zegt Falcke.

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    $\begingroup$ Are you sure about the "matter revolves around the hole in two days" part ? I've heard the event horizon has roughly a diameter of 100 billions km, and even at speed close to c that's about 2 weeks for a complete revolution $\endgroup$
    – Keelhaul
    Commented Apr 11, 2019 at 12:38
  • $\begingroup$ I'm just quoting one of the people involved in the project, I haven't verified his statements ... $\endgroup$
    – Glorfindel
    Commented Apr 11, 2019 at 12:39
  • $\begingroup$ The Schwarzchild radius of the M87 black hole is about 20 billion km. $\endgroup$ Commented Apr 12, 2019 at 12:27

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